Problem: Evaluate the triple integral. $ \int_0^1 \int_{z^2}^1 \int_z^x z - 4y \, dy \, dx \, dz =$ Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{-19}{70}$ (Choice B) B $\dfrac{-29}{70}$ (Choice C) C $\dfrac{-23}{70}$ (Choice D) D $\dfrac{-11}{70}$
Solution: We can evaluate triple integrals by repeated integration: $ \int_{a_0}^{a_1} \int_{b_0}^{b_1} \int_{c_0}^{c_1} f(x, y, z) \, dx \, dy \, dz = \int_{a_0}^{a_1} \left( \int_{b_0}^{b_1} \left[ \int_{c_0}^{c_1} f(x, y, z) \, dx \right] dy \right) dz$ The first layer: $\begin{aligned} &\int_0^1 \int_{z^2}^1 \int_z^x z - 4y \, dy \, dx \, dz \\ \\ &= \int_0^1 \int_{z^2}^1 \left[ zy - 2y^2 \right]_z^x dx \, dz \\ \\ &= \int_0^1 \int_{z^2}^1 zx - 2x^2 - (z^2 - 2z^2) \, dx \, dz \\ \\ &= \int_0^1 \int_{z^2}^1 zx - 2x^2 + z^2 \, dx \, dz \end{aligned}$ The second layer: $\begin{aligned} &\int_0^1 \int_{z^2}^1 zx - 2x^2 + z^2 \, dx \, dz \\ \\ &= \int_0^1 \left[ \dfrac{zx^2}{2} - \dfrac{2x^3}{3} + z^2x \right]_{z^2}^1 dz \\ \\ &= \int_0^1 \left( \dfrac{z}{2} - \dfrac{2}{3} + z^2 \right) - \left( \dfrac{z^5}{2} - \dfrac{2z^6}{3} + z^4 \right) dz \\ \\ &= \int_0^1 \dfrac{2z^6}{3} - \dfrac{z^5}{2} - z^4 + z^2 + \dfrac{z}{2} - \dfrac{2}{3} \, dz \end{aligned}$ The third layer: $\begin{aligned} &\int_0^1 \dfrac{2z^6}{3} - \dfrac{z^5}{2} - z^4 + z^2 + \dfrac{z}{2} - \dfrac{2}{3} \, dz \\ \\ &= \left[ \dfrac{2z^7}{21} - \dfrac{z^6}{12} - \dfrac{z^5}{5} + \dfrac{z^3}{3} + \dfrac{z^2}{4} - \dfrac{3z}{2} \right]_0^1 \\ \\ &= \dfrac{2}{21} - \dfrac{1}{12} - \dfrac{1}{5} + \dfrac{1}{3} + \dfrac{1}{4} - \dfrac{2}{3} \\ \\ &= \dfrac{2}{21} - \dfrac{1}{6} - \dfrac{1}{5} \\ \\ &= \dfrac{2}{21} - \dfrac{11}{30} \\ \\ &= \dfrac{-19}{70} \end{aligned}$ In conclusion: $ \int_0^1 \int_{z^2}^1 \int_z^x z - 4y \, dy \, dx \, dz = \dfrac{-19}{70}$